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From: Androcles on 1 Apr 2010 13:49

"Uncle Ben" <ben(a)greenba.com> wrote in message
news:b7abc813-9900-460f-a16f-d3c8cff07f33(a)20g2000vbr.googlegroups.com...
On Mar 31, 10:45 pm, "Androcles" <Headmas...(a)Hogwarts.physics_x>
wrote:

(snip)
==============================================
Happy to oblige, Bonehead, always pleased to snip when requested.
Mission accomplished.
Anything else I can snip for you?




From: Androcles on 1 Apr 2010 13:57

"Uncle Ben" <ben(a)greenba.com> wrote in message
news:b7abc813-9900-460f-a16f-d3c8cff07f33(a)20g2000vbr.googlegroups.com...


Newton believed his laws to be in effect at all times, not just when
the stick is in certain positions.

The force on the stick (action) due to the earth's gravitational
attraction is constant.
=============================================
Yes, Bonehead, that is what I said in -g*(1-cos(phi)), the 1 is
a constant. Didn't you know 1 is a constant, Bonehead? Never
could read algebra, could you?

From: Ray Vickson on 1 Apr 2010 18:10
On Mar 31, 6:27 pm, "Existential Angst" <UNfit...(a)UNoptonline.net>
wrote:
> Awl --
>
> I've found plenty of standard stuff on physical pendulums (meter stick
> pendulum), where I = 1/12 MR^2, period, etc, but what I can't find is this:
>
> With a meter stick standing straight up and pivoted at the *bottom*, what is
> the final angular velocity as it hits the table?
>
> My dim recollection, and perhaps why I can't find the solution so quick, is
> that an elliptic integral is involved??
>
> The application is an interesting one:
> I would like to approximate the force one would have to generate from a
> pushup position sufficient to thrust one's self up to a standing position..
>
> If I had the required Vo, I could probably figger stuff out from there.
>
> Any links, hints -- or solutions -- appreciated.
> --
> EA

Using the Lagrangian Lag = K - V, K = 1/2 * I * w^2, (I = 1/3 * m*
L^2) and V = m*g* L/2 * cos(u), u = u(t) = angle from vertical and w =
du/dt, the Euler-Lagrange equation is d/dt(@Lag/@w) = @Lag/@u, or 1/3
* m*L^3 * d^2 u(t)/dt^2 = 1/2 * m*g*L*sin(u(t)). Here is the Maple 11
solution (with u(0)=0, w(0) = a):
u(t) = solution of equation F(u) = t, where (or of F(u) = -t), where
F(z) = -2*(a^2*L+6*g)*L*(-g/a^2/L)^(1/2)*(g/(a^2*L
+6*g))^(1/2)*(sin(z)*((3*g+a^2*L-3*cos(z)*g)/(a^2*L+6*g))^(1/2)*(1-
cos(z))^(1/2)*(1+cos(z))^(1/2)*EllipticF(((3*g+a^2*L-3*cos(z)*g)/(a^2*L
+6*g))^(1/2),((a^2*L+6*g)/a^2/L)^(1/2))*(a^2*L^2)^(1/2)-(a^2*L/(a^2*L
+6*g))^(1/2)*EllipticF((a^2*L/(a^2*L+6*g))^(1/2),((a^2*L+6*g)/a^2/
L)^(1/2))*(L*(3*g+a^2*L-3*cos(z)*g))^(1/2)+(a^2*L/(a^2*L
+6*g))^(1/2)*EllipticF((a^2*L/(a^2*L+6*g))^(1/2),((a^2*L+6*g)/a^2/
L)^(1/2))*(L*(3*g+a^2*L-3*cos(z)*g))^(1/2)*cos(z)^2)/(L*(3*g
+a^2*L-3*cos(z)*g))^(1/2)/(a^2*L^2)^(1/2)/(-1+cos(z)^2)/g .

Good luck with that. Numerical methods are better in this case.

R.G. Vickson
From: Tim Little on 2 Apr 2010 01:48
On 2010-04-01, Existential Angst <UNfitcat(a)UNoptonline.net> wrote:
> The application is an interesting one: I would like to approximate
> the force one would have to generate from a pushup position
> sufficient to thrust one's self up to a standing position.

Under the usual physics simplifications it looks like a very
straightforward application of conservation of energy. No elliptic
integrals required.


- Tim
From: jbriggs444 on 2 Apr 2010 08:52
On Apr 1, 1:57 pm, "Androcles" <Headmas...(a)Hogwarts.physics_x> wrote:
> "Uncle Ben" <b...(a)greenba.com> wrote in message
>
> news:b7abc813-9900-460f-a16f-d3c8cff07f33(a)20g2000vbr.googlegroups.com...
>
> Newton believed his laws to be in effect at all times, not just when
> the stick is in certain positions.
>
> The force on the stick (action) due to the earth's gravitational
> attraction is constant.
> =============================================
> Yes, Bonehead, that is what I said in -g*(1-cos(phi)), the 1 is
> a constant. Didn't you know 1 is a constant, Bonehead? Never
> could read algebra, could you?

The force on the stick due to earth's gravitational attraction is
constant. Yes.
And by Newton's third law, the force on the earth due to the stick's
gravitational attraction is also constant.
So what?

You've made an assertion about the force of the pivot on the stick.

When the stick is vertical, the stick has zero acceleration. Newton's
_second_ law asserts that this means that the net force
on the stick is zero. That, in turn, means that the force of the
pivot on the stick is g. And this means that g(1-cos(phi)) matches
physical reality.

I respect your choice to factor out m.

So far, so good. Your formula fits reality in this boundary case.

Why should we expect this formula to continue to work?
Why should we expect zero vertical force at the pivot when the stick
is horizontal?

Let's examine that boundary case. *Flash of clarity -- gotta love it
when that happens*...

One should expect that vertical force to be _non-zero_ Here's the
argument...

Let's look at the stick when it is horizontal. We have the pivot over
on the left (in my mind's eye -- you may have it on the right).
The stick is in motion. It is falling at some speed. And it is
subject to the downward force of gravity. Not only is it falling, it
is _accelerating_ downward.

Its left end is held motionless by the pivot. Not only is the stick
falling, it is rotating.

At this instant, the rotation rate is _accelerating_ proportional to
the stick's vertical acceleration. It has to, because the pivot point
is level with the stick.

Shift viewpoint to an axis at the center of the stick. The stick has
rotational acceleration around this axis. It must, therefore, have a
torque about this axis. That torque can only come from the pivot
point. So the pivot point exerts a non-zero vertical force on the
stick.

QED. Your formula says zero. Reality says non-zero.
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